Relationship between effusion rate and molar mass

Graham's law of diffusion (video) | Khan Academy

relationship between effusion rate and molar mass

Although diffusion and effusion rates both depend on the molar mass of the gas . The small difference in molecular weights between UF6 and UF6 only . Effusion - The rate at which a gas escapes through a pinhole into a vacuum. Therefore, the density of a gas is directly proportional to its molar mass (MM). obtain the following relationship between the ratio of the velocities of the gases and. Molar mass is just one of the factors affecting effusion. Graham's law describes the relation of the rate of effusion with only one of the factors.

Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. How many effusion steps are needed to obtain Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity.

Use a logarithmic expression to compute the number of separation steps required.

relationship between effusion rate and molar mass

Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: Their atomic masses are 3. Helium-3 has unique physical properties and is used in the study of ultralow temperatures.

How many effusion steps are necessary to yield If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules.

Graham's law of diffusion

Because gas molecules collide as often as times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line. Summary Gaseous particles are in constant random motion. Gaseous particles tend to undergo diffusion because they have kinetic energy.

relationship between effusion rate and molar mass

Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. Effusion refers to the movement of gas particles through a small hole. And I'm drawing them as O22 molecules, or two atoms, rather, of oxygen in one molecule of O2.

And I've also got some carbon dioxide here. And of course, carbon dioxide-- in the name you can already hear it-- it's got dioxide. So, it's got two oxygens as well. It almost looks like a little bow tie. So, I've got a few of these molecules.

So, I've got equal proportions of both. So, I have a friend over here, an alien friend, and I asked my friend to stand some distance away from my pot.

And you can see I'm actually cooking my pot. I've got some fire with firewood underneath it. And I say, please stand 10 feet away from my pot, sir. And this little alien friend is a good friend of mine, so he says, no problem.

And the reason I'm asking him to help me with it is because he has a very special nose, a very, very special nose. He's never in his a life smelled oxygen or carbon dioxide.

He's lived on Planet Graham his whole life. And Planet Graham has these little green molecules. But he has such a special nose that he can actually detect whether he's smelling carbon dioxide or oxygen. So, I'm going to actually take this lid off. And I'm going to say, hey, if you detect with your special nose either one of these-- let's say that these molecules, one of them goes over and kind of goes into his nose-- if you can detect it, please let me know which one you're smelling.

Graham's law - Wikipedia

And that's my test. And I want to know which of these molecules, oxygen or carbon dioxide, is going to reach his nose, which is 10 feet away, first.

So it's basically a race. And you can make a prediction right now as to which molecule you think is going to get to his nose first, the oxygen or the carbon dioxide. Now you might think, oh, it's very easy. There's a direct path. But actually remember, these molecules, these green molecules in the planet atmosphere, are whizzing around.

They're going in all sorts of different directions. And as a result, they're going to smack into our carbon dioxide or oxygen molecules as they try to make their way over there. And kind of a random fact, but an interesting one to think about, is that in our atmosphere we have a lot of nitrogen, a lot of nitrogen gas.

Now, if you took one nitrogen gas molecule, which is N2, and let it go, and measured its speed and kind of clocked it, it would be going at about 1, miles an hour. But the only reason it doesn't actually go that speed in reality is because the molecules of nitrogen, they actually will smack into each other and bounce off of each other millions and millions of times every second. And so because they're smacking and colliding constantly, they never really reach those real potential speeds.

They go much slower. So really what we're talking about is when molecules are bouncing and clanging into each other and slowly making progress towards our little alien's nose, that is the idea of diffusion. They're going to kind of rattle around and slowly make their way over to his nose. And maybe if I came back, let's say, 10 minutes later, maybe this little oxygen would be right here.

Maybe you might have a little carbon dioxide right here. Slowly making progress towards the nose. That's what we're trying to figure out-- which one will get over there first.

Graham's Laws of Diffusion and Effusion - Chemistry LibreTexts

So, you've had time to think about it. And I'm actually going to tell you how I think we should approach the problem, which is thinking back to kinetic energy.

relationship between effusion rate and molar mass

We're heating this thing up, so we're putting thermal, or heat energy, into the molecules. Both types of molecules are getting the same amount. I've got the oxygen getting some kinetic energy. I'm going to put a little o for oxygen.

relationship between effusion rate and molar mass

And it's going to equal, or should equal, the amount of energy that my carbon dioxide is getting. And I'm going to do that as a little c for carbon dioxide. So these two molecule types should be getting the same amount of energy. Now remember, it's not like it's one molecule we're thinking of. We're thinking of many, many molecules. So first, I'm going to have to change these units a little bit.

2.9: Graham's Laws of Diffusion and Effusion

Because again, I'm thinking about the individual molecule. So I've got to figure out what these molecules weigh. And v is going to change over to rate, or diffusion rate. And the reason I'm doing that is because, again, I'm thinking about the overall diffusion of the gas. It's not like I'm betting on any one molecule. I'm betting on the entire population of carbon dioxide molecules beating out the population of oxygen molecules, or vice versa, the oxygen molecules beating out the carbon dioxide molecules.

But not an individual molecule. So I have to think of the average rate that those molecules are moving. So, let me rewrite this equation.

relationship between effusion rate and molar mass